Question: What is the extraneous solution to these equations? $\dfrac{x^2}{x - 5} = \dfrac{4x - 4}{x - 5}$
Explanation: Multiply both sides by $x - 5$ $ \dfrac{x^2}{x - 5} (x - 5) = \dfrac{4x - 4}{x - 5} (x - 5)$ $ x^2 = 4x - 4$ Subtract $4x - 4$ from both sides: $ x^2 - (4x - 4) = 4x - 4 - (4x - 4)$ $ x^2 - 4x + 4 = 0$ Factor the expression: $ (x - 2)(x - 2) = 0$ Therefore $x = 2$ The original expression is defined at $x = 2$ and $x = 2$, so there are no extraneous solutions.